# G Chem question..calculate molality

#### TeethRCool

##### Member
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5+ Year Member
What is the molality of 85% phosphoric acid, H3PO4, if the density of the solution is 1.70 g/ml?

Answer: 1000*(1/1.70)*(100.0 / 15.0)*(1.70/1)*(85.0 / 100.0)*(1/98) m

Someone care to explain? Thanks so much!

#### dat_student

##### Junior Member
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5+ Year Member
TeethRCool said:
What is the molality of 85% phosphoric acid, H3PO4, if the density of the solution is 1.70 g/ml?

Answer: 1000*(1/1.70)*(100.0 / 15.0)*(1.70/1)*(85.0 / 100.0)*(1/98) m

Someone care to explain? Thanks so much!
molality = The molal concentration of a solute, usually expressed as the number of moles of solute per 1,000 grams of solvent (or 1 kg)

molality = moles of H3PO3 / kg of solvent

H3PO4 = 3 * 1 + 31 + 16 * 4 = 98 grams/mole

85% = 85 grams H3PO4
the rest (15%) = 15 grams solvent

(1)
85 grams H3PO4 * (1/98 moles/gram) = # of moles of solute

(2)
15 grams solvent = 15 * (10^-3) kg (per 1 kg of solvent)

85 grams H3PO4 * (1/98 moles/gram)......1000 * 85
________________________________ = ___________ m
15 * (10^-3) kg...................................15 * 98

that's the simple form of the answer (everything else cancels out)

1000*(1/1.70)*(100.0 / 15.0)*(1.70/1)*(85.0 / 100.0)*(1/98) m

#### ttran01

##### Member
10+ Year Member
it's only been a semester since ive taken gen chem 1.....i already forgot everything